Question

A ray of light strikes the midpoint of one face of an equiangular (60°–60°–60°) glass prism (n = 1.5) at an angle of incidence of 40.9°. (a) Trace the path of the light ray through the glass, and find the angles of incidence and refraction at each surface. (b) If a small fraction of light is also reflected at each surface, find the angles of reflection at the surfaces. Step 1 The angles of incidence ??? from air to glass and ??? from glass to air, and the angles of refraction ??? in the glass and ??? in the air as the light ray exits the prism are shown in the diagram. 60° ? ? ??? ??? ??? ??? Air Glass (a) The angle of incidence at the first air–glass surface is ??? = 40.9°. We know that the index of refraction of air is n_air = 1.0 and the index of refraction of this glass n_glass = 1.5. Snell's law gives the angle of refraction at the first air–glass surface, where sin ??? = sin 40.9° = , as ??? = sin?¹(n_air sin ??? / n_glass) = sin?¹[ 1.0 ( ) / 1.5 ] = °.

          A ray of light strikes the midpoint of one face of an equiangular (60°–60°–60°) glass prism (n = 1.5) at an angle of incidence of 40.9°.
(a) Trace the path of the light ray through the glass, and find the angles of incidence and refraction at each surface.
(b) If a small fraction of light is also reflected at each surface, find the angles of reflection at the surfaces.

Step 1
The angles of incidence ??? from air to glass and ??? from glass to air, and the angles of refraction ??? in the glass and ??? in the air as the light ray exits the prism are shown in the diagram.

60°
? ?
??? ??? ??? ???
Air Glass

(a) The angle of incidence at the first air–glass surface is ??? = 40.9°. We know that the index of refraction of air is n_air = 1.0 and the index of refraction of this glass n_glass = 1.5. Snell's law gives the angle of refraction at the first air–glass surface, where sin ??? = sin 40.9° = , as
??? = sin?¹(n_air sin ??? / n_glass) = sin?¹[ 1.0 ( ) / 1.5 ] = °.

$A ray of light strikes the midpoint of one face of an equiangular (60°–60°–60°) glass prism (n = 1.5) at an angle of incidence of 40.9°. (a) Trace the path of the light ray through the glass, and find the angles of incidence and refraction at each surface. (b) If a small fraction of light is also reflected at each surface, find the angles of reflection at the surfaces. Step 1 The angles of incidence ??? from air to glass and ??? from glass to air, and the angles of refraction ??? in the glass and ??? in the air as the light ray exits the prism are shown in the diagram. 60° ? ? ??? ??? ??? ??? Air Glass (a) The angle of incidence at the first air–glass surface is ??? = 40.9°. We know that the index of refraction of air is nair = 1.0 and the index of refraction of this glass nglass = 1.5. Snell's law gives the angle of refraction at the first air–glass surface, where sin ??? = sin 40.9° = , as ??? = sin?¹(nair sin ??? / nglass) = sin?¹[ 1.0 ( ) / 1.5 ] = °.$

Added by Lindsey W.

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